27r~-~ Coefficient estimates for negative powers of the derivative of univalent functions

Ark. Mat., 36 (1998), 255-273

@ 1998 by Institut Mittag-Leffier. All rights reserved

Coefficient estimates for negative powers of the derivative of univalent functions

D a n i e l B e r t i l s s o n

A b s t r a c t . Let f be a one-to-one analytic function in the unit disc with f ' ( 0 ) = l . We prove sharp estimates for certain Taylor coefficients of the functions (if)P, where p<0. The proof is similar to de Branges' proof of Bieberbach's conjecture, and thus relies on LSwner's equation. A special case leads to a generalization of the usual estimate for the Schwarzian derivative of f. We use this to improve known estimates for integral means of the functions

If'l p

for integers p<_ 2.

1. I n t r o d u c t i o n a n d r e s u l t s

L e t f b e a u n i v a l e n t (i.e., o n e - t o - o n e a n a l y t i c ) f u n c t i o n in t h e u n i t disc ]z t < 1 w i t h f ( 0 ) = 0 a n d f ' ( 0 ) = l . L e t S d e n o t e t h e set of such f . W e e q u i p S w i t h t h e t o p o l o g y of l o c a l l y u n i f o r m convergence; t h u s S is c o m p a c t [6, p. 9]. L e t p b e a r e a l n u m b e r , a n d c o n s i d e r t h e coefficients of t h e f u n c t i o n

(1.1)

(f'(z))P=~'~Cn,pZ ~,

w h e r e C 0 , p = l .

n ~ 0

T h e c o n t i n u o u s f u n c t i o n a l f H ICn,pl a s s u m e s a m a x i m u m on S. T h e p r o b l e m of de- t e r m i n i n g (or e s t i m a t i n g ) t h i s m a x i m u m is a n i n t e r e s t i n g p r o b l e m , w h i c h is r e l a t e d t o e s t i m a t e s for t h e i n t e g r a l m e a n s of t h e f u n c t i o n

]f,12p.

T h i s follows f r o m t h e i d e n t i t y

(1.2)

2~T O(2

I f ' ( r e ~ ~

27r~-~ _{I I.Qp.}

² r ^{2 n} .

n ~ O

F o r i n s t a n c e , B r e n n a n ' s c o n j e c t u r e [3]

27r

fo If'(rei~ dO=O((1-r)-l-e)

for a l l c > 0

2 5 6 D a n i e l B e r t i l s s o n

is equivalent to

N

(1.3) Z ]c~,-112 = O(NI+~) for all c > 0.

n : 0

Our main result concerns negative p.

T h e o r e m 1.

Let c~,p be the nth coefficient of

( f t ) p , as

in

(1.1).

Assume that p<O and that

1 4 n < - 2 p + 1 .

Then the maximum of ]Cn,pl over f E S is attained

when f is the Koebe .function f ( z ) = z / ( l +z) 2.

In other words, if f cS, then

(1.4) ]c,,,pl <_ Cn,p,

where C~,p is defined by

(1.5) ( l ~ z ) 3 =

Ck,pz k, where

Co,; = 1.

(Ck,p>O for all k>_O,

/ f p < - ~ . )

Moreover, equality is attained in

(1.4)

if and only if

Z 2, where j l=l.

Our proof of Theorem 1 is parallel to de Branges' proof of Milin's conjecture [5], [7]: We consider a L6wner family of single-slit mappings

ft.

LSwner's equation

t p

leads to a linear system of differential equations for the coefficients of (f~) . In this way the problem of maximizing ]cn,p] becomes a problem in optimal control theory. This problem is solved by proving that a certain quadratic expression in the coefficients is an increasing fimction of t. The computations in this last step are more involved t h a n in de Branges' proof. In de Branges ~ proof this was easy, since one could use known inequalities for Jacobi polynomials.

Unfortunately, it is not true for every

f E S

that

]cn,_~l<_Cn, 1

for all n. (This would imply Brennan's conjecture (1.3).) An example is given by the function

f ( z ) = e x p - 1 . 2 6 7 4~ d(,

k = 0

which has

c~,p#O(n ~176

if p _ < - l , see [8, proof of Corollary 3]. On the other hand, Cauchy's formula for the coefficients of the derivative of (1.5) gives

(1.6)

C~,p_:O(n,p] 1)

i f p < 0 .

Coefficient e s t i m a t e s for negative p o w e r s of the derivative of univalent functions 257

Thus, if - 1 . 0 6 4 < p _ < - 1 , then there is a function

f E S

such that l e n , p l > C n , p for infinitely many n. This also holds if - l < p < 0 , since then there is a function

f E S

with

log

If'(re i~ 12p dO

(1.7) lim inf > max{21p I - 1, 0},

see [4, p. 34] and [13, T h e o r e m 8.6]. (If we had

IC~,pl<lCn,pl

for all n, then (1.6) and (1.2) would give a contradiction to (1.7).)

The case n - - - 2 p + l of Theorem 1 turns out to be especially interesting. Con- sider the differential operator

T h e o r e m 1 states that (1.8)

S n f = ( f ' ) ( n 1)/2Dn(f, )

(n-1)/2.

ISnf(o)l <n!Cn,

(n-l)/2.

1 times the Sehwarzian derivative. Like the Schwarzian The operator $2 is just - ~

derivative, Sn has an invariance property,

(1.9)

Sn(fo~-)

^-- ( ( S n I ) ~ n if 7- is a M6bius transformation.

See [9] for a general discussion of operators with this property. Using a disc auto- morphism T we can thus "move" the estimate (1.8) to an arbitrary point of the unit disc.

T h e o r e m 2.

For functions f univalent in the unit disc we have the sharp estimate

ISnf(Z)l= (f'(z)) (''-1)/2

< K n ( 1 IZl2) - n ,

where K n = ( n - 1 ) ( n + l)(n+ 3) ...

( 3 n - 3 )

and n is a positive integer.

When n = 2 this is the usual estimate for the Schwarzian derivative [6, p. 263].

Using this estimate for the Schwarzian derivative, Pommerenke [13, T h e o r e m 8.5]

proved that

ff

(1.10)

II'(re~~ 1dO=O((1-r)-~176

i f f E S .

We use Pommerenke's argument and Theorem 2 to prove the following estimates for integral means.

258 Daniel Bertilsson

T h e o r e m 3.

Let En be the positive root of

E ( E + 1)(E+2)... ( E + 2 n - 1) = K~, where Ks is as in Theorem 2, and

n > l

is an integer.

If f is a univalent function in the unit disc, then

fo ^{If'(rei~ dO} = o((1 for

^all

> ^O.

In particular,

~o 2~ If'(re i~

1-2

dO

= O((1 - r ) - 1.547),

fro 2~ lf'(re~~ dO

= O((1 -r)-2"53~

These estimates are small improvements of the known estimates

f02~r ( ( 1 ~1p1--0.399~

I f ' ( r e i ~ ]

for p < - l ,

which follow from (1.10) and the elementary estimate

[f'(z) l> ~ if'(0)l(1-

Izl). Note t h a t as n ~ + o c we have

En=n-~-+o(1).

This should be compared with Carleson's and Makarov's result [4, Corollary 1]

I ^{If'(rei~ dO}

^{= O} for large negative p.

T h e exponent IPl- 1 is best possible (take f as a Koebe function).

In the next two sections we prove T h e o r e m 1, except for the characterization of the extremal functions, which will be proved in Section 4. We prove Theorems 2 and 3 in Sections 5 and 6, respectively.

2. T h e p r o o f o f T h e o r e m 1

Let F: [0, + o c ) - - ~ C b e a p a r a m e t r i z a t i o n of a J o r d a n arc such that, for s o m e T > 0 , the arc F([T, +e~)) is an interval [F(T), +cx~) on the positive real axis. Assume that f0 ~ S maps onto the complement of the arc F([0, +oc)). The set of mappings of the type f0 is dense in S [6, p. 81]. Thus it is sufficient to prove T h e o r e m I for

f = f o .

For t > 0 , let

ft

be the Riemann mapping of the unit disc 'onto the complement of

Coefficient e s t i m a t e s for n e g a t i v e p o w e r s of t h e d e r i v a t i v e of u n i v a l e n t f u n c t i o n s 259

the arc C([t, +c~)), normalized so that f t ( 0 ) = 0 and f / ( 0 ) > 0 . We can choose the parametrization F so that

f[(O) e ~.

LSwner's differential equation [1, Chapter 6]

then relates the mappings ft,

, l+w(t)z

(2.1)

? e ( z ) = f ~ ( z ) z l _ w ~

for t > 0 ,

where w is a complex-valued continuous function with I~o(t)l= 1, and the dot denotes differentiation with respect to t.

Fix an integer n and a negative number p such that l < n < - 2 p + l . We are interested in the coefficients of the functions

o ~

g t ( z ) = ( e - t f [ ( z ) ) V = E c k ( t ) z k ,

where

co(t) 1.

k--O

Specifically, our task is to prove the inequality

(2.2) Ic,~(0)l < Cn,

where

C~=Cn,p

is defined by (1.5). T h a t C n > 0 will be proved after Lemma 4 in Section 3. To prove (2.2) we use two facts. First, the function e

T f T E S

maps onto the complement of an interval

[x,

+oc) on the positive real axis. Thus e TfT(z )=

z/(l + z) 2,

which implies

ck(T) Ck fork_>0.

Second, differentiation of LSwner's equation (2.1) gives

, , l+co(t)z (12aa(t)z-co(t)~ (1-co(t)z)

22c~

)

= ,

which leads to a linear system of differential equations for the coefficients of

gt,

k 1

(2.3)

d k ( t ) - - k c k ( t ) + E ( 2 j + 2 v ( k - j + l ) ) w ( t ) k - J c j ( t )

for k > 0 .

j - - 0

We can write this in matrix form using the vectors

c(t)=(co(t),cl(t),...,c,~(t)) T

and

C=(Co,CI,...,Cn) T

and the lower triangular matrix M(c0) with elements

(2j+2p(k-j+l))co k J

f o r 0 < j < k < _ n ;

M(w)kj = k for

O<_j=k<_n;

0 forO<_k<j<_n.

We have thus reduced our task to the following control theory problem.

2 6 0 D a n i e l B e r t i l s s o n

C o n t r o l p r o b l e m .

Let c(t) be the solution of the initial value problem

(2.4) d(t) =

M(co(t))c(t), 0 < t < T, c(T) = C,

where w(t) is a continuous function of modulus 1. Prove that len(O)l_<C,~.

Note that the choice co(t)--1 corresponds to

ft(z) etzl(l+z) 2

and c n ( t ) = C~. Following de Branges, we solve the above problem by studying the quadratic expression

T~

(2.5) H(t) = Z hk (t) Ick (t) l 2,

k = O

where the weights

hk(t)

are real-valued and depend on t, but not on the function co.

We get

n

(t) = ~ h~ (t) I ck (t) l ~ + h~ (t) 2 ae(c~ (t) +k (t))

k - - O

(2.6)

)

= h ~ ( t ) l ~ k ( t ) 1 2 + h K t ) 2 R e M ( 1 ) k j T j ( t ) ,

k 0 j=0

where we have introduced

~yj(t)=co(t)-Jcj(t)

and used equation (2.3). Using the diagonal matrix

D(t)

with diagonal elements h0(t), ...,

h~(t)

and the vector ~/(t)=

(70(t), ..., 7,,~(t)) r , we c a n write this as

f~I(t) = ~ { ) T (D(t) + M(1)T D(t) + D(t)M(1) )7(t).

We now t r y to determine the functions hk so that the following conditions are satisfied:

(2.7) /:/(t) >_ 0 for t > 0 and for all choices of w, (2.8) /:/(t) = 0 if w = 1,

(2.9) h k ( 0 ) = 0 for k < n and h ~ ( 0 ) = l .

If these conditions are satisfied, our problem is solved, since we get [cn(0)12=H(0)_<

~" 2

H(T) ~ k o hk(T)C~,

with equality if c o ( t ) - l ; hence ten(0)[_<Cn.

When establishing (2.7), we will forget that the 7k(t) are not arbitrary, and prove the stronger statement that the matrix

(2.10)

P(t) 19(t)+M(1)TD(t)§

is positive semi-definite for t_>0.

Coefficient e s t i m a t e s for negative powers of t h e derivative of univalent functions 261

Let us first show that conditions (2.8) (2.10) determine hk uniquely. In the case w(t)--i we have 7 ( t ) = C , so condition (2.8)gives

c T p ( t ) C = O .

Condition (2.10) now yields X / ~ C = 0 , whence

P(t)C = D(t)C + M(1)T D(t)C + D ( t ) M ( 1 ) C = O.

However, M ( 1 ) C = 0 (the case a~(t)-I of equation (2.4)). Thus we get

(2.11) ~)(t) =

--M(1)Ty(t),

where

y(t) = D ( t ) C = (Yo (t), ... , y~ (t) ) T.

Condition (2.9) implies the initial condition

(2.12) y(o) = (o, o, ..., o, Cn) T.

Thus

(2.13) h ~ ( t ) = s ~ ( t ) / C k ,

k=0,1,...,~,

are uniquely determined.

Now, define

y(t)

and

hk(t)

by equations (2.11) (2.13). Then we have

P(t)C=O,

so condition (2.8) is satisfied, as well as (2.9). It remains to prove condition (2.10), t h a t is, t h a t the Hermitian form r~(t)=-yTP(f)7, also given by the rigtxt-hand side of (2.6), is positive semi-definite.

From now on we consider 3'k as fi'ee variables (independent of t and c~). The quantity r](t) takes a simple form expressed in the new variables

k

c~k = E ( k - j + l ) ~ / j , j 0

k = 0 , 1 , . . . , n .

We can write 7k=ctk--2ak_l + a k - 2 and

k

j = 0

where c~ 1 =OZ 2 = 0 . Substituting this into (2.6) and collecting terms one gets (2.14)

n

v(t) = y~ u~ (t)I~ ? +vk (t)2 ae ( ~ ~ : ~ ) + w ~ (t)2 R e ( ~ ) ,

k=O

262 Daniel Bertilsson

where

(2.15)

(2.16) (2.17)

uk (t) = hk (t) +4hk+l(t) +hk+2 (t) +2khk

(t)

+ ( - 16p+8)hk+l (t) + ( - 4 p - 4 - 2k)hk+2 (t),

vk (t) = - 2 h k (t) - 2hk+l (t) + (419- 2 - 2k)hk (t) + (8/)+ 2k)hk+l (t),

wk(t) h~(t)- 2phk(t),

and

hn+l(t)=hn+2(t)=O.

We want to complete the squares and write

n

( 2 . 1 8 ) ~](t) =

Z p k ( t ) lozk 4-qk(t)o~k l+rk(t)o~k_2t 2.

k=0

First, we prove t h a t this is possible if t is large.

L e m m a 1.

If

n < - 2 p + l

then

n

(2.19)

h k ( t ) = Z h k j e - J t > O fort>O,

k = 0 , 1 , . . . , n ,

j k

where hkj are constants and

hkk>0.

Proof.

By (2.13) and (2.11) we have

(2.20)

hk(t)+khk(t)=C~ 1 ~ ( - 2 k - 2 p ( j - h + l ) ) C j h j ( t ) , j-k+1

where the coefficient of

hi(t)

is positive thanks to the assumption n < 2 p + l . A descending induction on k, using also the initial condition (2.9), now proves (2.19).

Moreover,

hkk--hk(O)+ f o ekt(hk(t)+khk(t))dt>O. []

By continuity, we can henceforth assume t h a t n < - 2 p + l . L e m m a 1 implies t h a t

uk(t), vk(t)

and

wk(t)

are polynomials in e t

uk(t) = khkke-kt+... , vk(t)

= (4p--2)hkke kt+... ,

wk(t)-- (-2p-k)hkke ~t+... ,

where the o m i t t e d t e r m s are of smaller order as t--~+oc. This shows t h a t for large t we can successively complete squares in (2.14), beginning with t e r m s containing an.

In each step of this process the coefficient of t e r m s of t y p e lakl 2 will have leading

Coefficient estimates for negative powers of the derivative of univalent functions 263 term

khkke kt.

Thus we get (2.18) with

pk(t)

> 0 for k = l , 2, ..., n, for large t. Since

P(t)

is a singular matrix, r/(t) is also singular, so we must have p 0 ( t ) = 0 for large t.

The functions pk(t),

qk(t), rk(t),

now defined for large t, are rational functions of e t. We now want to prove that these functions have no poles for 0<e-t_<1, and that

pk(t)>O

for t_>0, k = l , 2, ..., n. Thanks to the special structure of r/(t), we can derive a one-step recursion formuIa for Pk (t). Identifying coefficients in (2.14) and (2.18) we get, for large t,

(2.21)

u k ( t ) = p k ( t ) § 2,

O<k<n,

(2.22)

vk(t) = p k ( t ) q k ( t ) + p k + s ( t ) q k + l ( t ) r k + s ( t ) , 1 < k < n,

(2.23)

wk (t) = Pk(t)rk (t), 2 < k < n,

where

Pn+l Pn+2 =qn+l ~rn+l

= r n + 2 ~ 0 .

~n t h e case "r C we have

~(~)=CrP(t)C=O,

so (2.18) t o g e t h e r w i t h p ~ ( t ) > O for l < k < n and

po(t)

0 gives, for large t,

(2.24)

Ak+qk(t)Ak-l+rk(t)Ak-2 O, l < k < n ,

where

Ak=Ey=o(k--j+l)Cj.

Solve (2.24) for q~(t), plug this into (2.22) and use (2.23). This gives the recursion formula

A k ~ Ak

1Ak+l Ak-1

(2.25)

pk(t)-- Ak wk(t)-- A~ wk+l(t)- Ak vk(t) A~A 21wk+l(t)

2pk+l(t) for l < k < n - 1 , but still only for large t. In the next section we use this formula to prove the following lemma.

L e m m a 2.

The meromorphic functions Pk ( t ) are regular for t >_ O, and for t > 0 we have

(2.26)

pn(t)=nhn(t)>O, po(t)=O,

(2.27)

p k ( t ) > A ~ w k ( t ) + ~ 2 ( k + l ) h k ( t ) > _ O f o r k = l , 2 , . . . , n - 1 .

By (2.23) and (2.24) the functions rk and qk are also regular for t > 0 . Hence (2.18) holds for t > 0 , and thus r/(t)_>0.

Remark.

If n > - 2 p + l and p < - } the proof breaks down: We get u n - l ( t ) < 0 for large t; hence

P(t)

is not positive semi-definite.

264 Daniel Bertilsson

Remark. Casting de Branges' proof of Milin's conjecture into our notation we note the following differences. Corresponding to (2.14) one gets the simpler expres- sion

n

?(t) -- ~ ~ (t)la~ }2 + ~ (t)2 Re(a~a~_l).

k--0

This stems from the fact that one studies the functions 0t(z)=log(e-tft(z)/z), which contain no derivative of ft. The simpler structure of ~(t) implies that the functions /Sk(t) corresponding to our functions Pk (t) become polynomials in e -e.

3. T h e p r o o f o f L e m m a 2

For convenience, introduce the parameter q = - 2 p + l > n . We need some lem- mas concerning the constants Ck, Ak and B k = A k - A k - l = C o + . . . + C k .

L e m m a 3. The following recursion formulas hold for all integers k:

(3.1) kC~ = (2q-2)c~_1 -(q-k+l)Ck_2,

(3.2) kBk : ( 2 q - 1 ) B k - l - ( q - k ) B k 2,

(3.3)

kAk = 2qAk 1 - ( q - k - 1 ) A k - 2 . Here, Ck-~Bk = A k = 0 for k<0.

These formulas follow from applying (1-z2)d/dz on the generating Proof.

functions

C O

(1-

~)~(1+~) - ~ : ~

c ~ ~,

k=O

(I - ~)~- ~ (1 + z)-~ = ~ s ~ ~,

k = O

(1-z)'-2(l+z) -3~:~Akz< []

k=0

L e m m a 4. If q>3, then kCk>qCk-l>O for l < k < q + l .

Proof. The case k : l is clear. Inductively, assume that kCk>qCk 1>0, where l<k<q. Together with equation (3.1) this implies

( k + l ) C k + l >_

(2q-2)Ck-(q-k)kCk/q>qCk. []

This shows that Ck>0 if 4 < k < q + l . Moreover, for q>5, Co=1, C 1 = 2 q - 2 , C2=(q- 1)(2q- 5) and Ca= 89 1)(4q 2-11q+9) are all positive. Equation (3.1) now implies that Ck>0 for all k_>0, if q>5.

Coefficient estimates for negative powers of the derivative of univalent functions 265

L e m m a 5. / f q > 3 ,

then

C k _ l / C k < C k / C k + 1

for

0 < k < q .

Proof.

T h e case k = 0 is trivial. Inductively, a s s u m e t h a t

Ck Ck 2 < C~_

1, where l < k < q . E q u a t i o n (3.1) implies

(lg-~-J_)Ck+lCk_l-lgC 2 (q-1-1--k)CkCk_ 2 ( q - k ) C 2 1 .

B u t L e m m a 4 implies Ck >

Ck

2, so we get

(k-~-])(Ck+lCs 1 - C 2 ) < ( q lg)(CkCk_2-C 2 1 ) < 0 . [~]

L e m m a 6.

If

q>3,

then Ak-1/Ak<Bk-1/Bk for

l < k < q .

Proof.

B y L e m m a 5 we have

B j _ I _ c 0 + . . . + C j _ l <

Co+...+cj _ Bj

B5 Co+...+C~ Co+...+C5+1 Bj+I

^for

1-<j -<q.

T h u s

Ak-~ Bo+...+Bk 1 Bk 1

-- < - - for l < k < q .

A~ Bo+...+Bk Bk []

L e m m a 7.

If

q > 3

and l<k<q, then

(3.4) C k B k + l 1

Ck+lBk < 14 q_~.

Proof.

T h e case k = 1 is easily checked. Inductively, assume t h a t

Ck_iBk 1

- - < 1 - t

CkBk_l q + l - k '

where

2<_k<q.

Using

C k = B k - B k - 1

a n d the recursion formula (3.2) we can write this as

(3..5)

- k ( q - k + l ) B ~ + ( 2 q ( q - k ) + 2 q - 1 ) B k B k _ l - ( q k+2)(q-k)B~ 1

> 0 . We want t o prove t h e inequality (3.4), which in a similar w a y c a n be w r i t t e n

(3.6)

(2q-l-(k+l)(q-k+l))B~+((2q-2)(q-k)+l)BkBk

¹

- ( q - k - 1 ) ( q - k ) B ~ 1 > 0 .

266 Daniel Bertilsson

It suffices to prove t h a t the difference between the l e • h a n d sides of (3.6) and (3.5),

= ( B k - B k _ l ) ( ( q + k - 2 ) B k - 3 ( q - k ) B k _ l )

is positive. Lemmas 5 and 4 imply

Bk Co +... + Ck Ck q

Bk-~ C o + . . . + G - ~ > Ck-~ > k"

Thus we need only prove that ( q + k - 2 ) q / k - 3 ( q - k ) > 0 , which is a simple verifica- tion. [5

We also need the following facts. By (2.20), (2.19) and h ~ ( t ) = e ~t we have hk (t)+ k hk (1)> 0 if t > 0 and 0 < k < n - 1 , and so equation (2.17) and Lemma 1 yields (3.7) w~(t) > 0 f o r t > O , i f O < k < n - 1 .

Together with Lemma 1, this proves the second inequality of (2.27). By (2.16) and (2.17) we have

vk (t) = --2wk (t) -- 2wk+l (t) -- 2 (k + 1)hk (t) -- 2 ( q - k - 1)hk+l (t), which substituted into (2.25) gives the recursion formula

(3.s)

Ak-1

pk(t)= ~ [(2 --Ak-2• Ak_l j~wk(~5)@ (2-- A~: 1)wk+l(ti)@2(k@l)hk(t)

Ak--lw~+l(t)z] l < k < n - 1 . + 2 ( q - k - 1 ) h k + l ( t ) dkpk+l(t) J'

Now (2.26) follows from (2.21), (2.15) and (2.20),

(3.9) p,,~ (t) = un(t) = h ~ ( t ) + 2 n h ~ ( t ) = nh~(t) = ne -~t > 0 We prove inequality (2.27) by descending induction over k.

I n d u c t i o n b a s e . For t>O and 2<nKq,

An 3 . . A~ 2 -

n--1 n 1

Proof. By (3.8) and (3.9), this is equivalent to

(3.10) 2 ( l - ~ ) w ~ _ l ( t ) + ( 2 - A ~ _ l A ~ _ (t) + 2 ( q - n)h (t) /

for t > O.

An 2 w~(t) 2 An 1 nhn(t) ^{> 0 .}

Coefficient e s t i m a t e s for negative powers of t h e derivative of univalent functions 267

By (2.20) and (2.17) we have

wn l(t) = (q-n)(2CnC~l_lhn(t)+hn-l(t))

and

wn(t) = (q-n-1)hn(f),

which substituted into

(3.10)

gives

an_2j(q-n)h'*-l(t) + ~ n _ 2 ) ( q - n ) ~ _ l +

+2(q-n)-AnAn-12 (q n_-l)21h~(t)>O

Since

An-a<A~ 2

and hn l ( t ) > 0 , we only have to prove that the coefficient of

hn(t)

is positive. Using the recursion formula (3.3) with

k=n,

we can write this coefficient as

V{1A,,

(3.11)

4(q-n) L\ An-2] Cn 1 1 q--1

5 that It follows from L e m m a 4 and

C2/C1 = q - a

C~ q 1

> - -

Cn-1 2n

L e m m a 5 implies that

A~ 3 Cn_3+2Cn_4+...+(n-2)Co Cn 1

An-2 - Cn_2+ 2Cn_3+... + ( n - 2)C1-}-(n-1)Co < -~-n

Thus

(3.11)

is positive. []

I n d u c t i o n s t e p . Assume that l < k < n - 2 , t>_0 and

> +

r ~ k + l

Hence, by (3.7) and L e m m a 1,

Ak-1 . ,

(3.12) pk+l(t) >

~k+lwk+l(t) > O.

We want to prove that

pk(t)>(Ak-2/Ak)wk(t)+(Ak-1/Ak)2(k+l)hk(t).

By the recursion formula (3.8) and (3.12) it is enough to prove

268 D a n i e l B e r t i l s s o n

Using the functions s k ( t ) = E j n _ k ( j - - k + l ) y j ( t ) we can write the differential equa- tion (2.11) as

(3.14) ~lk (t) = --ksk(t) § 2(q-- 1)Sk+l (t) + (--q§ l+k)sk+2 (t).

Remember that h k ( t ) = y k ( t ) / C ~ > O . Substituting this, y k ( t ) = s k ( t ) - - 2 s k + l ( t ) + sk+2(t) and (3.14) into (2.17) we get

lo k (~) = C k I ((q-- 1 - k)sk (t) + ksk+2 (t)).

Putting this into (3.13) and using hk+l(t)>_0, we see t h a t it suffices to prove (3.15) ( q - l - k ) s k ( t ) + k s k + 2 ( t ) - # k ( q 2 - - k ) s k + l ( t ) - - p k ( k + l ) s k + 3 ( t ) > O, where

J~k =

Ck ( ~ k + l 1)

Ck+l _ CkBk+lAk 1 > O.

1 - A k - ~ Ck+lBk 1A~

Ak-1

Note t h a t q > n > 3 . From Lemma 6 and Lemma 7 it follows t h a t

CkBk+l BkAk 1 1

= - - < 1 + - - .

~k Ck+lBk Bk 1Ak q - k

Hence

8 k (15) __ Yk (t) ~- 2yk+ 1 (t) @... ~- ( n - - ~-~ 1)y n (t) > 11, -- k -~1 8 k + 1 ( ~ ) y k + l ( t ) + . . . + ( n - - k ) y n ( t ) - n - k ^{> Pk,} and similarly sk+2 (t) >#ksk+3 (t). Thus

( q - 1 - - k ) ( s k ( t ) - - ~ k S k + l ( t ) ) ~ k ( S k + 2 ( t ) - - t l k S k + 3 ( ] ~ ) ) > O, which implies (3.15), since 8k4_ 1 (t) • 8k+ 3 (t).

4. T h e e x t r e m a l f u n c t i o n s

Clearly, the Koebe functions f ( z ) = z / ( l + A z ) 2, where

I l:l,

give equality in Theorem 1. Assume that f ( z ) = z + a 2 z 2 + a a z a + . . . E S is not a Koebe function. As in [7] we use an approximation argument to prove that this implies strict inequality in (1.4). The function f is the limit of a sequence of functions f~'~ of the type f0 considered in Section 2. We denote the corresponding quantities with the same

Coefficient e s t i m a t e s for n e g a t i v e p o w e r s of t h e d e r i v a t i v e of u n i v a l e n t f u n c t i o n s 269

symbol as in Section 2, but with a superscript m. Assume first t h a t l < n < q = - 2 p + 1. Since c~ n (0) =

2pa~ ~,

Bieberbach's theorem l a2l < 2 [12, T h e o r e m 1.5] implies that

I C ~ ( 0 ) I < 4 1 p l - 5 = C ~ ~ f o r , ~ > , ~ o , where 5>0. T h e differential equation (2.3) implies

Ida's(t)[ =

IcF(t)+4pa3(t)l <_

Cl@4lpl.

Thus

I c F ( t ) [ < C 1 - 5 f o r m > m 0 , 0 < t < t 0 ,

where 5 > 0 and t 0 > 0 are independent of m. Now equation (2.18) gives, for

m>mo

and 0<t<t0,

[l"~(t) >_ pt (t)]7?(t)+ 27~(t)+q1(t)7~(t)l 2 =pl (t)lw(t)-lc?(t)-Cll 2

> p l ( t ) 5 2 . By L e m m a 2 we have

pl(t)>_hl(t)/q,

so we get

fro ~2 fotO $2

Ic~(O)l 2 = gin(O) ~ ~ hk(T)C~- qhl(t) dt = C ~ - qhl(t) dr.

k = 0

In the limit m---,oc this shows that

IC~,p[<Cn.

By the proof of L e m m a 1, this conclusion also holds in the limit case

q=n,

except when n = 2 , in which case h l ( t ) ~

3 2

0. In this exceptional case we have

c2,-1/2

~ ( a 2 - a a ). Thus it follows from the elementary estimate l a ~ - a 3 1 < l [12, Theorem 1.5] that we have strict inequality in (1.4). This concludes the proof of T h e o r e m 1.

Remark.

Just like our estimate of c2,_1/2, the estimate of e3, 1 is not new. It was proved by Ozawa [11] in the form [a4

3a2a3+2a~l<2

using Schiffer's varia- tional method.

5. G e n e r a l i z e d S c h w a r z i a n d e r i v a t i v e s Define the differential operator Sn for conformal maps f by

S n f = (f')(n-1)/2D~(f')

(n-1)/2,

where we use the same branch of vzf 7 at both occurrences. To prove the invariance property (1.9) we use the following lemma [2], [10].

270 D a n i e l B e r t i l s s o n

L e m m a 8.

Let T be a MSbius transformation. If two analytic functions are related by [l-(go~-)(r')-(n-~)/~, then their nth derivatives have the relation

~(~)=

( g ( n ) O T ) ( T t ) ( n 4 - 1 ) / 2 .

Proof.

Since r is a composition of transformations of the types

z~--~z+a, z ~ b z

and

z~--, 1/z,

it suffices to prove the lemma when ~- is one of these. T h e first two cases are rather trivial. In the third case

T(z)=l/z,

we can by continuity and linearity assume t h a t

g(z)=z k,

and then an easy calculation proves the lemma. []

Now let

g=(f,)-(n-1)/2

and

~]--((for)') -(n-1)/2.

Since

g=(gor)(r') -(n-1)/2,

the lemma shows t h a t

(5.1)

S~(for)- ~(~) _g(~)or

got (r') ~ = ( ( S n f ) o7)(7') ~,

i f T i s a MSbius transformation.

We now prove Theorem 2. T h e o r e m 1 gives

(5.2)

[S~f(O)l<~n!Cn, (~

1)/2

if

f c S .

Since S~ is homogeneous this holds also if f is just univalent in the unit disc. By (1.5), Cn,-(~-l)/2 is the n t h coefficient of

(l__z)_(n

1)/2(1+z)3(n_1)/2

₌ --(n 1)/2 ( _ z ) k E 3(n 1)/2 _{Z 2 ,}

k = 0 j 0

so that

~-~(-(n--kl)/2)(3(~--12/2)(_1)k Kn

C n ' - ( n - 1 ) / 2 = n ! '

k = 0

where

K~=(n-1)(n+l)...

( 3 n - 3 ) Now let

r(z)=(z+C)/(l+Cz),

where 1r and let f be a univalent function in the unit disc. T h e n

for

is also univalent in the unit disc, so (5.2) gives ]Sn(for)(0)l_<Kn. Together with the invariance property (5.1), this shows that

I S n f ( r = I S ~ f ( r ( 0 ) ) l = s ~ ( f o ~ ) ( 0 ) < K n ( < ( 0 ) ) n - (1_1r since T'(0)=I--[~[2.

Coefficient estimates for negative powers of the derivative of univalent functions 271 6. E s t i m a t e s for i n t e g r a l m e a n s

We now use T h e o r e m 2, following P o m m e r e n k e [13, p. 181] to prove T h e o r e m 3.

L e m m a 9. If g is analytic in the unit disc and m ( r ) = J : ~ [g(rei~ 2 dO, then

/oo

rn(2~)(r) <_4 ~

Ig(n)(rei~ dO.

o o ]~ o o

Writing g(z)=}-~k= 0 bkz , we get m ( r ) = 2 7 r ~ k = 0 Ibkl 2r2k" T h e l e m m a Pro@

is evident from a comparison of coefficients in

and

o o

m(=')(~) = 2~ ~ Ib~l~2k(2k-1)...

( 2 k - 2 n § 2k-2n

k n

Ig(~)(<~{~ ~ dO = 2 ~ ~ i b ~ k ( k - 1 ) . . . ( k - n + l ) 1 2 r 2k 2n. []

k = n

Let n > 1 be an integer and let f be a univalent function in the unit disc. Using L e m m a 9 with g=(ff)-(n-1)/2 we get

f0

m (2n) (r) << 4 ~ I g ( r P ~ 1 7 6 dO.

T h e o r e m 2 now gives the differential inequality

K~ ~2 Ig(rei~

^{dO <} l ~ 2 n m ( r )

_ _

( _ )

for r0_<r< 1. T h e corresponding differential equation

m(2n)(r)=\l~ro]

^{(1 r) 2n}

has solutions ~ ( r ) = C ( i - r ) E(,o), where E(ro) is the positive solution of

( 2 ~2n 2

E ( E + I ) ... ( E + 2 n - 1 ) = \ l ~ r 0 / Kn"

Choosing C large enough, we get

m(k)(ro)<Cn(k)(ro), k = 0 , 1 , . . . , 2 n - - 1 ,

272 Daniel Bertilsson and so Proposition 8.7 of [13] gives

(6.1) m ( r ) < ~ ( r ) f o r r o _ < r < l .

A n o t h e r proof of (6.1). T h e function A ( r ) m ( r ) - C n ( r ) satisfies

( 2 } 2 A(r) r0_<r<l,

(6.2) A(2n)(r) _< ~ ( l _ r ) ~ ,

and A ( k ) ( r 0 ) < 0 for k = 0 , 1 , . . . , 2 n . Let r l < l be the largest number such t h a t A(2*~)(r)<0 for r o < r < r l . If r l < l , then A ( 2 n ) ( r l ) = 0 and A ( r l ) < 0 , which con- tradicts (6.2). Thus r 1 = 1 , and (6.1) follows.

We thus have

~0

^2~r

r e ( r ) = I f ' ( r e i ~ r ) - E ( r ~ Since E(ro)--*E~ as r0---~l, T h e o r e m 3 is proved.

R e f e r e n c e s

1. AHLFORS, L. V., Conformal Invariants: Topics in Geometric Function Theory, Mc- Graw-Hill, New York, 1973.

2. BOL, G., Invarianten linearer Differentialgleichungen, Abh. Math. Sem. Univ. Ham- burg 16 (1949), 1 28.

3. BRENNAN, J. E., The integrability of the derivative in conformal mapping, J. London Math. Soc. 8 (1978), 261 272.

4. CARLESON~ L. and MAKAROV, N. G., SoIile results connected with Brennan's con- jecture, Ark. Mat. 32 (1994), 33 62.

5. DE BRANGES, L., A proof of the Bieberbach conjecture, Acta Math. 154 (1985), 137-152.

6. DUREN, P. L., Univalent Functions, Springer-Verlag, New York, 1983.

7. FITZGERALD, C. H. and POMMERENKE, C., The de Branges theorem on univalent functions, Trans. Amer. Math. Soc. 290 (1985), 683-690.

8. GNUSCHKE-HAUSCHILD, D. and POMMERENKE, C., On Bloch functions and gap series, J. Reine Angew. Math. 367 (1986), 172 186.

9. GUSTAFSSON, B. and PEETRE, J., M6hius invariant operators on Riemann surfaces, in Function Spaces, Differential Operators and Nonlinear Analysis (Sodankylii, 1988) (P/iiv/~rinta, L., ed.), Pitman Res. Notes Math. Set. 211, pp. 14 75, Longman, Harlow, 1989.

10. GUSTAFSSON~ B. and PEETRE, J., Notes on projective structures on complex mani- folds, Nagoya Math. J. 116 (1989), 63 88.

11. OZAWA, M., On certain coefficient inequalities of univalent functions, Khdai Math.

Sem. Rep. 16 (1964), 183 188.

Coefficient estimates for negative powers of the derivative of univalent functions 273

12. POMMERENKE, C., Univalent Functions, Vandenhoeck & Ruprecht, G6ttingen, 1975.

13. POMMERENKE, C.~ Boundary Behaviour of Conformal Maps, Springer-Verlag, Berlin Heidelberg, 1992.

Received August 3, 1997 Daniel Bertilsson

Department of Mathematics Royal Institute of Technology S-100 44 Stockholm

Sweden

email: [email protected]